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2 years ago

Stephen gathered data about the average rainfall in two cities. He organized his data in the showing graph.

Mathematics

1 answer:

PSYCHO15rus [73]2 years ago

5 0

Answer: B and R

Step-by-step explanation:

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A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman

Yanka [14]

Answer:

$W\approx 8.72$ and $L\approx 15.57$ .

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

$2L+W+\frac{1}{2}(2\pi r)=38$

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

$2L+W+\frac{1}{2}(2r\pi)=38$

$2L+W+\frac{\pi}{2}W=38$

Let us find area of window equation as:

$\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)$

$\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)$

$\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})$

$\text{Area}=W\cdot L+\frac{\pi}{8}W^2$

Now, we will solve for L is terms W from perimeter equation as:

$L=38-(W+\frac{\pi }{2}W)$

Substitute this value in area equation:

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

Since we need the area of window to maximize, so we need to optimize area equation.

$A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2$

$A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2$

Let us find derivative of area equation as:

$A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W$

$A'=38-2W-\pi W+\frac{\pi}{4}W$

$A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W$

$A'=38-2W-\frac{3\pi W}{4}$

To find maxima, we will equate first derivative equal to 0 as:

$38-2W-\frac{3\pi W}{4}=0$

$-2W-\frac{3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}=-38$

$\frac{-8W-3\pi W}{4}*4=-38*4$

$-8W-3\pi W=-152$

$8W+3\pi W=152$

$W(8+3\pi)=152$

$W=\frac{152}{8+3\pi}$

$W=8.723210$

$W\approx 8.72$

Upon substituting $W=8.723210$ in equation $L=38-(W+\frac{\pi }{2}W)$ , we will get:

$L=38-(8.723210+\frac{\pi }{2}8.723210)$

$L=38-(8.723210+\frac{8.723210\pi }{2})$

$L=38-(8.723210+\frac{27.40477245}{2})$

$L=38-(8.723210+13.70238622)$

$L=38-(22.42559622)$

$L=15.57440378$

$L\approx 15.57$

Therefore, the dimensions of the window that will maximize the area would be $W\approx 8.72$ and $L\approx 15.57$ .

8 0

3 years ago

An Italian restaurant made 6 breadsticks for a table of four people. How many breadsticks will they need to make if they are exp

inessss [21]

Answer:

i think that is 147

Step-by-step explanation:

6 0

2 years ago

What is the answer to the whole thing? I'm having trouble with my homework.

notsponge [240]

Y=8,4,2,-2 linear and in that order

8 0

3 years ago

Read 2 more answers

Given the original number n. Multiply the number by 9. Add 99. Divide this sum by 9. Subtract the original number, n, from the q

tangare [24]

Answer:

Step-by-step explanation:

Hello,

Given the original number n.

$\large \boxed{n}$

Multiply the number by 9.

$\large \boxed{9n}$

Add 99.

$\large \boxed{9n+99}$

Divide this sum by 9.

$\large \boxed{\dfrac{9n+99}{9}=n+11}$

Subtract the original number, n, from the quotient.

$\large \boxed{n+11-n=11}$

Thank you.

4 0

2 years ago

* Let S = Span {(2,-1, 1), (3, 1, 1), (1, 2, 0)}. (i) Calculate the dimension of S.

Sholpan [36]

The span of 3 vectors can have dimension at most 3, so 9 is certainly not correct.

Check whether the 3 vectors are linearly independent. If they are not, then there is some choice of scalars $c_1,c_2,c_3$ (not all zero) such that

$c_1 (2,-1,1) + c_2 (3,1,1) + c_3 (1,2,0) = (0,0,0)$

which leads to the system of linear equations,

$\begin{cases} 2c_1 + 3c_2 + c_3 = 0 \\ -c_1 + c_2 + 2c_3 = 0 \\ c_1 + c_2 = 0 \end{cases}$

From the third equation, we have $c_1=-c_2$ , and substituting this into the second equation gives

$-c_1 + c_2 + 2c_3 = 2c_2 + 2c_3 = 0 \implies c_2 + c_3 = 0 \implies c_2 = -c_3$

and in turn, $c_1=c_3$ . Substituting these into the first equation gives

$2c_1 + 3c_2 + c_3 = 2c_3 - 3c_3 + c_3 = 0 \implies 0=0$

which tells us that any value of $c_3$ will work. If $c_3 = t$ , then $c_1=t$ and $c_2 = -t$ . Therefore the 3 vectors are not linearly independent, so their span cannot have dimension 3.

Repeating the calculations above while taking only 2 of the given vectors at a time, we see that they are pairwise linearly independent, so the span of each pair has dimension 2. This means the span of all 3 vectors taken at once must be 2.

3 0

1 year ago